Question

There are four +ive numbers $G_1,G_2,G_3,G_4$ in G.P. which satisfy the following conditions (i)$G_2^2+G_3^2=250$ (ii)$G_1{G}_4=\alpha$where$\alpha$ is greater root of the equation $x^{2+lo{g}_{{10}^x}}=(0.001{)}^{-8/3}$.FInd $G3/G2$ ?(given $G1,G2.G3,G4$ are in increasing order)

Answer 1

$G_1=a,G_2=ar,G_3=a{r}^2,G_4=a{r}^3$

$G_2^2+G_3^2=250$

$a^2{r}^2(1+r)=\mathrm{250....}\left(1\right)$

$x^{2+\log x}=(0.001{)}^{-8/3}$

Applying logarithm on both sides

$(\log x+2)\log x=8⟹\log x=-4,2⟹\alpha =100$

$G_1{G}_4=a^2{r}^3=\mathrm{100....}\left(2\right)$

$\frac{\left(1\right)}{\left(2\right)}⟹\frac{1+r}{r}=\frac{5}{2}⟹r=\frac{2}{3}$

$G_1=(\frac{2}{3}{)}^{3}a,G_2=(\frac{2}{3}{)}^{2}a,G_3=\frac{2}{3}a,G_4=a$ $(\because G_1,G_2,G_3,G_4$ are in increasing order $)$

$\frac{G_3}{G_2}=\frac{3}{2}$